Compute the sum $ \sum_1^{\infty} \ln{ \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} }$

Question

I need to prove that the sum $ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) $ converges and equals $ \ln(\frac{4}{3})$. I tried expanding the fraction into four terms, which cancel each other out, but I'm left with $\ln(4)$. Here's my attempt:

$ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) = \sum_{1}^{\infty} \ln{(n+1)} - \sum_{1}^{\infty} \ln{n} + \sum_{1}^{\infty} \ln{(3n+1)} -\sum_{1}^{\infty} \ln{(3n+4)} \\ = \sum_{1}^{\infty} \ln{(n+1)} - \sum_{2}^{\infty} \ln{n} + \ln(1) + \sum_2^{\infty} \ln{(3n+1)} + \ln(4) -\sum_{1}^{\infty} \ln{(3n+4)} = \ln(4)$

The last step is simply about playing correctly with the indices and the lower bounds of the series. These telescopic series almost completely cancel each other out, though I can't find the right answer.

Please tell me what I missed. Thanks

Answer 1

You have to be very careful with the telescoping series in this exercise. For example, for each $k \in \mathbb{N}$:

$$\sum_{n=1}^k \ln \left(\frac{n+1}{n}\right) \ = \ \ln\left(\prod_{n=1}^k \frac{n+1}{n}\right)$$ $$=\ln(k+1),$$ so taking limits w.r.t $k$, this diverges.

What is going on here. In general, let $\{a_n\}; n \in \mathbb{N}$ be a sequence of real numbers. Then: $$\sum_{n=1}^{k} a_n - \sum_{n=1}^{k} a_{n+1} =a_1-a_{k+1},$$ so a limit as $k \rightarrow \infty$ exists if and only if $\lim_{k \rightarrow \infty} a_k = 0$.

In any event, note that for each $k \in \mathbb{N}$:

$$\sum_{n=1}^k \ln \left(\frac{3n+1}{3(n+1)+3}\right) \ = \ \ln \left(\frac{4}{3(k+1) +3}\right),$$ whereas

$$\sum_{n=1}^k \ln \left(\frac{n+1}{n}\right) = \ln(k+1).$$

Can you take it from here. In particular, you want to evaluate $$\lim_{k \rightarrow \infty} \ln \left( \frac{4}{3(k+1)+3}\right)+\ln(k+1)$$ $$ = \lim_{k \rightarrow \infty} \ln \left(\frac{4(k+1)}{3(k+1)+3}\right).$$

Answer 2

The issue with your calculation is that the sums are not individually convergent. Therefore, you cannot write

$$\sum_{n=1}^\infty \log \frac{3n+1}{3n+4} = \sum_{n=1}^\infty \log (3n+1) - \sum_{n=1}^\infty \log(3n+4),$$ because each of those sums on the RHS are infinite.

Instead, you need to look at the limiting behavior for a corresponding finite partial sum. Define

$$f(m) = \sum_{n=1}^m \log \frac{3n+1}{3n+4}.$$ Then $f(m)$ is finite for any positive integer $m$, and now we have no problem computing the telescoping sum:

$$\begin{align} f(m) &= \sum_{n=1}^m \log (3n+1) - \sum_{n=1}^m \log (3(n+1) + 1) \\ &= \sum_{n=1}^m \log (3n+1) - \sum_{n=2}^{m+1} \log (3n+1) \\ &= \log 4 - \log (3(m+1)+1) \\ &= \log \frac{4}{3m+4}. \end{align}$$

Similarly, $$g(m) = \sum_{n=1}^m \log \frac{n+1}{n} = \log (m+1) - \log 1 = \log (m+1).$$

Consequently, $$f(m) + g(m) = \log \frac{4(m+1)}{3m+4} = \log \frac{4 + 4/m}{3 + 4/m},$$ for all positive integers $m$,

and now taking the limit as $m \to \infty$, we get

$$\sum_{n=1}^\infty \log \frac{(n+1)(3n+1)}{n(3n+4)} = \lim_{m \to \infty} f(m) + g(m) = \lim_{m \to \infty} \log \frac{4 + 4/m}{3 + 4/m} = \log \frac{4}{3}.$$

Answer 3

Let's do it carefully:

$\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$

$\ln(\prod_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$

$\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})$.

Now $\prod_{n=1}^k\frac {n+1}n$ clearly equals $k+1$ as it's just the telescoping series $\require{cancel} \frac {\cancel 2}1\frac {\cancel 3}{\cancel 2}\frac {\cancel4}{\cancel3}.... \frac {\cancel{k-1}}{\cancel{k-2}}\frac {\cancel k}{\cancel{k-1}}\frac {k+1}{\cancel k}$.

[But note $\prod_{n=1}^\infty \frac {n+1}n = \lim_{k\to \infty}(k+1) = \infty$ diverges. As others have pointed out you can't do an $\lim M\cdot N= \lim N\cdot \lim N = \infty \cdot 0 = ?????$ split which is what you are in essence trying to do.]

$\prod_{n=1}^k \frac {3n+1}{3n+4}= \frac {3n+1}{3(n+1)+1}$ is also a telescopic series. And its product will be $\frac {4}{3(k+1)+1}$ as it is the telescopic seriers $\frac {4}{\cancel 7}\frac{\cancel 7}{\cancel {11}}......\frac {\cancel {3(k-1)+1}}{\cancel{3k+1}}\frac{\cancel{3k+1}}{3(k+1)+1}=\frac 4{3k+4}$.

[Also note $\prod_{n=1}^{\infty} \frac {3n+1}{3n+4} =\lim_{k\to \infty} \frac 4{3k+4} = 0$. So if we tried to do a $\lim \prod \frac {(n+1)(3n+1)}{n(3n+4)} = \lim (n+1)\frac 4{3k+4} = \lim (n+1)\cdot \lim \frac 4{3k+4}=\infty \cdot 0 = ?????$ it will not work. That is just illegal.]

So $\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})=\ln((k+1)\frac 4{3k+4})=\ln \frac{4(k+1)}{3(k+1)+1}=\ln \frac 4{3+\frac 1{k+1}}$

[But this is acceptable. If you have an $\lim MN = \lim M \cdot \lim N = \infty \cdot 0$ paradox, it might be fixable if the product $MN$ have terms that cancel out as it does in this case. $\lim (k+1)\frac 4{3(k+1)+1}\ne \lim(k+1)\lim\frac 4{3(k+1)+1} = \infty \cdot 0=???$ is not okay but $\lim (k+1)\frac 4{3(k+1)+ 1} = \lim \cancel{(k+1)}\frac 4{3\cancel{(k+1)} + \frac 1{\color{purple}{k+1}}}=\frac 43$ is perfectly legal.]

And that's pretty much it:

$\ln(\sum_{n=1}^\infty \frac {(n+1)(3n+1)}{n(3n+4)})=$

$\ln(\lim_{k\to \infty}\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$

$\lim_{k\to \infty}\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$ (whoa! careful! is that actually true? [yes, it is..... so long as the sum converges... which it must else the logs of the partial sums would not converge which we know they do. ])

$\lim_{k\to \infty}\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})=$

$\lim_{k\to \infty}\ln \frac {4}{3+\frac 1{k+1}}=$

$\ln (\lim_{k\to \infty} \frac 4{3+\frac 1{k+1}}=$

$\ln \frac 43$